## Trigonometry (11th Edition) Clone

$$\cos\theta=\frac{2\sqrt5}{5}$$ $$\sin\theta=\frac{\sqrt5}{5}$$
$$\cos2\theta=\frac{3}{5} \hspace{1.5cm}\theta\hspace{0.2cm}\text{terminates in quadrant I}$$ $$\sin\theta=?\hspace{2cm}\cos\theta=?$$ 1) First, we decode the signs of the $\sin\theta$ and $\cos\theta$ As $\theta$ terminates in quadrant I, it can be stated that $\sin\theta\gt0$ and $\cos\theta\gt0$. 2) Using only the Double-Angle Identities for $\cos2\theta$, we can find out the value of $\sin\theta$ and $\cos\theta$ as follows. $\cos2\theta$ can be written as $$\cos2\theta=2\cos^2\theta-1$$ Thus, $$\cos^2\theta=\frac{\cos2\theta+1}{2}=\frac{\frac{3}{5}+1}{2}=\frac{\frac{8}{5}}{2}=\frac{4}{5}$$ $$\cos\theta=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\hspace{1cm}\cos\theta\gt0$$ Now, $\cos2\theta$ can also be written this way $$\cos2\theta=1-2\sin^2\theta$$ Thus, $$\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-\frac{3}{5}}{2}=\frac{\frac{2}{5}}{2}=\frac{1}{5}$$ $$\sin\theta=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}\hspace{1cm}\sin\theta\gt0$$